The sum of two positive numbers is 20% of the sum of their squares and 25% of the difference of their squares. If the numbers are x and y then `(X+Y)/(X^(2))` equal to

Options

A. `1/4`
B. `3/8`
C. `1/3`
D. `2/9`

Answer

To solve the problem step by step, we will begin by setting up the equations based on the information provided in the question.

### Step 1: Set up the equations
We know that:
1. The sum of two positive numbers \( x \) and \( y \) is 20% of the sum of their squares.
2. The sum of \( x \) and \( y \) is also 25% of the difference of their squares.

From the first statement, we can write:
\[ x + y = 0.2(x^2 + y^2) \]
This can be rewritten as:
\[ x + y = \frac{1}{5}(x^2 + y^2) \quad \text{(Equation 1)} \]

From the second statement, we can write:
\[ x + y = 0.25(x^2 - y^2) \]
This can be rewritten as:
\[ x + y = \frac{1}{4}(x^2 - y^2) \quad \text{(Equation 2)} \]

### Step 2: Set the two equations equal to each other
Since both equations equal \( x + y \), we can set them equal to each other:
\[ \frac{1}{5}(x^2 + y^2) = \frac{1}{4}(x^2 - y^2) \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 4(x^2 + y^2) = 5(x^2 - y^2) \]

### Step 4: Expand and rearrange the equation
Expanding both sides:
\[ 4x^2 + 4y^2 = 5x^2 - 5y^2 \]
Rearranging gives:
\[ 4y^2 + 5y^2 = 5x^2 - 4x^2 \]
This simplifies to:
\[ 9y^2 = x^2 \quad \text{(Equation 3)} \]

### Step 5: Substitute \( x^2 \) in terms of \( y^2 \)
From Equation 3, we can express \( x^2 \) as:
\[ x^2 = 9y^2 \]

### Step 6: Find \( x + y \)
Substituting \( x^2 \) back into Equation 1 to find \( x + y \):
\[ x + y = \frac{1}{5}(x^2 + y^2) \]
Substituting \( x^2 = 9y^2 \):
\[ x + y = \frac{1}{5}(9y^2 + y^2) = \frac{1}{5}(10y^2) = 2y^2 \]

### Step 7: Find \( \frac{x+y}{x^2} \)
Now we need to find \( \frac{x+y}{x^2} \):
\[ \frac{x+y}{x^2} = \frac{2y^2}{9y^2} \]
This simplifies to:
\[ \frac{x+y}{x^2} = \frac{2}{9} \]

### Final Answer
Thus, the value of \( \frac{x+y}{x^2} \) is:
\[ \frac{2}{9} \]